Math-like chaining of the comparison operator - as in, "if ( (5<j<=1) )" [duplicate]

int j=42;
if( (5<j<=1) ) {
    printf("yes");
} else {
    printf("no");
}

Output:

yes

Why does it output yes?
Isn't the condition only half true?

C does not understand math-like syntax, so

if(1<j<=5)

is not interpreted as you expect and want; it should be

if (1 < j && j <= 5)

or similar.

As explained in other answers, the expression is evaluated as

 ((1 < j) <= 5)

 =>  ("true" <= 5)

 =>  "true"

where "true" (boolean value) is implicitly converted to 1, as explaneid e.g. here, with references to standards too, and this explain why "true" has to be "less than" 5 (though in C might not be totally correct to speak about "implicit conversion from bool to int")

As per operator precedence and LR associativity,

1<j evaluates to 1

1<=5 evaluates to 1

if(1)
{ 
  printf("yes")